Then work on the second term in that expansion to pull out the factor of "1". What you're left with at that point will of course be a "proper" rational fraction, with the degree of the numerator lower than that of the denominator.

Thank you for the reply. Yes, and it's because I saw that I can't reduce that equation, in particular, to a 1st_order/1st_order+constant form (knowing my weak mathematical skills), that I thought there might be another way.

What I obtained (after scribbling a page of notebook) was s-1+2/(s+1), apparently it's a good result because wolframalpha showed the same. But, again, I thought (hoped) that it may be another way to allow for a translation in a state-space form. This was (is) for a custom block in LTspice to allow for 2nd order Laplace transfer functions with RLC (transient simulation is very bad and even un-recommended by the help file, ac is OK, though). Testing is done with a custom state-space block and I had problems implementing a random, but plausible, function to test it. So, I got here.

Thanks for the help, though, a doubt cleared means one way closed and another opened.

Thank you for the reply. Yes, and it's because I saw that I can't reduce that equation, in particular, to a 1st_order/1st_order+constant form (knowing my weak mathematical skills), that I thought there might be another way.

What I obtained (after scribbling a page of notebook) was s-1+2/(s+1), apparently it's a good result because wolframalpha showed the same. But, again, I thought (hoped) that it may be another way to allow for a translation in a state-space form. This was (is) for a custom block in LTspice to allow for 2nd order Laplace transfer functions with RLC (transient simulation is very bad and even un-recommended by the help file, ac is OK, though). Testing is done with a custom state-space block and I had problems implementing a random, but plausible, function to test it. So, I got here.

Thanks for the help, though, a doubt cleared means one way closed and another opened.

If I understood correctly, this is a rather important question as it involves partial fractions (of rational functions), something pretty important in integration.

In the present case, since the numerator's degree is higher than the denominator's , we can directly divide polynomialwise:

[tex]s^2+1=(as+b)(s+1)+r=as^2+(a+b)s+b+r\Longrightarrow a=1\,,\,a+b=0\Longrightarrow b=-1\,,\,b+r=1\Longrightarrow r=2[/tex]
Thus,
[tex]\,s^2+1=(s-1)(s+1)+2\Longrightarrow\frac{s^2}{s+1}=s-1+\frac{2}{s+1}[/tex]
Of course, this case is so simple that it can be done almost "by eye", but the general case can be way more involved. Google "Partial fraction"